package com.cg.leetcode;

import org.junit.Test;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

/**
 * 30.串联所有单词的子串
 *
 * @author cg
 * @program LeetCode->LeetCode_30
 * @create 2022-06-27 20:02
 **/
public class LeetCode_30 {

    @Test
    public void test30() {
        System.out.println(findSubstring("ababababab", new String[]{"ababa", "babab"}));
        System.out.println(findSubstring("barfoothefoobarman", new String[]{"bar", "foo"}));
        System.out.println(findSubstring("barfoofoobarthefoobarman", new String[]{"bar", "foo", "the"}));
        System.out.println(findSubstring("wordgoodgoodgoodbestword", new String[]{"word", "good", "best", "good"}));
        System.out.println(findSubstring("lingmindraboofooowingdingbarrwingmonkeypoundcake", new String[]{"fooo", "barr", "wing", "ding", "wing"}));
    }

    /**
     * 给定一个字符串 s 和一些 长度相同 的单词 words 。找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。
     * 注意子串要与 words 中的单词完全匹配，中间不能有其他字符 ，但不需要考虑 words 中单词串联的顺序。
     * <p>
     * 示例 1：
     * 输入：s = "barfoothefoobarman", words = ["foo","bar"]
     * 输出：[0,9]
     * 解释：
     * 从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。
     * 输出的顺序不重要, [9,0] 也是有效答案。
     * <p>
     * 示例 2：
     * 输入：s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
     * 输出：[]
     * <p>
     * 示例 3：
     * 输入：s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
     * 输出：[6,9,12]
     * <p>
     * 提示：
     * 1 <= s.length <= 104
     * s 由小写英文字母组成
     * 1 <= words.length <= 5000
     * 1 <= words[i].length <= 30
     * words[i] 由小写英文字母组成
     *
     * @param s
     * @param words
     * @return
     */
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<>();
        HashMap<String, Integer> need = new HashMap<>(), window;
        for (int i = 0; i < words.length; i++) {
            need.put(words[i], need.getOrDefault(words[i], 0) + 1);
        }
        int wordLength = words[0].length();
        int len = words.length;
        int count = len * wordLength;
        for (int i = 0; i <= s.length() - count; i++) {
            String sub = s.substring(i, i + count);
            window = new HashMap<>();
            for (int j = 0; j < sub.length(); j += wordLength) {
                String substring = sub.substring(j, j + wordLength);
                if (need.containsKey(substring)) {
                    window.put(substring, window.getOrDefault(substring, 0) + 1);
                }
            }
            if (need.equals(window)) {
                res.add(i);
            }
        }
        return res;
    }

}
